What is the slope of the line tangent to $f(x) = -2x^{2}+3x+6$ at $x = 2$ ?
Answer: The slope of the tangent line is $ \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{(-2(x+\Delta x)^{2}+3(x+\Delta x)+6) - (-2x^{2}+3x+6)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{(-2(x^{2}+2x \Delta x+\Delta x^{2})+3(x+\Delta x)+6) - (-2x^{2}+3x+6)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{-2x^{2}-4(x \Delta x)-2\Delta x^{2}+3x+3(\Delta x)+6+2x^{2}-3x-6}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{-4(x \Delta x)-2\Delta x^{2}+3(\Delta x)}{\Delta x}$ $ = \lim_{\Delta x \to 0} -4x-2(\Delta x)+3$ $ = -4x+3$ $ = (-4)(2)+3$ $ = -5$